How do you solve the triangle given a=345, b=648, c=442?

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How do you solve the triangle given a=345, b=648, c=442?

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Answer 1 · 2017-10-12T11:49:16

Solution of triangle: $a = 345, b = 648, c = 442, ∠ A \approx {29.97}^{0},$ $a=345,b=648, c=442, /_A~~ 29.97^0 ,$
$∠ B \approx {110.24}^{0}, ∠ C \approx {39.79}^{0}$ $/_B~~ 110.24^0 , /_C~~ 39.79^0$

Explanation:

Sides of triangle are $a = 345, b = 648, c = 442$ $a=345,b=648, c=442$

$cos (A) = \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{648^{2} + 442^{2} - 345^{2}}{2 \cdot 648 \cdot 442} = 0.8663$ $cos(A) = (b^2 + c^2 − a^2)/(2bc) = (648^2 + 442^2 − 345^2)/(2*648*442) = 0.8663$
$:. /_A=cos^-1(0.8663)~~ 29.97^0$

$cos(B) = (c^2 + a^2 − b^2)/(2ca) = (442^2 + 345^2 − 648^2)/(2*442*345) = -0.34597$
$:. /_B=cos^-1(-0.34597)~~ 110.24^0$

$/_C ~~180-( 29.97+110.24) ~~ 39.79^0$

Solution of triangle: $a=345,b=648, c=442, /_A~~ 29.97^0 ,$
$/_B~~ 110.24^0 , /_C~~ 39.79^0$ [Ans]

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How do you solve the triangle given a=345, b=648, c=442?

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